MaplePrimes Posts

MaplePrimes Posts are for sharing your experiences, techniques and opinions about Maple, MapleSim and related products, as well as general interests in math and computing.

Latest Post
  • Latest Posts Feed
  • What's up with Mapleprimes's timing today? Every time I click on a Post or Question, the main post loads at the normal speed (which is slow to start with), and then it takes a full minute for the Answers or Replies (or lack thereof) to load. This has been consistent for every Post or Question that I've read today. Is anyone else experiencing this?

    In a recent conversation I explained whyLSODE was giving wrong results (http://www.mapleprimes.com/questions/210948-Can-We-Trust-Maple#comment230167). After a lot of confusions and weird infinite loops for answers, it turned out that Newton Raphson was not properly done.

    Both LSODE and MEBDFI are currently incompletely implemented (only one iteration is done instead of Newton Raphson till convergence). Maplesoft should update the help files accordingly.

    The post below explains how better results are obtained with method = mgear. To run the command mgear you will need Maple 6 or earlier versions. For lsode, any current version is fine.  Unfortunately Maple deprecated an algorithm that worked fine. From Maple 8, the algorithm moved to Rosenbrock methods for stiff equations. This is still not ideal.

    If Maple had a working algorithm, I am hoping that Maplesoft folks would consider bringing it back in future versions. (At least with the same functionality as in Maple 6).

    PLEASE NOTE, the issue is not with solving this example (Very simple). This example is chosen to show how a popular algorithm in the literature is wrongly implemented.

     

    Here Maple's lsode is forced to take only one step and use first order back ward difference formula to integrate from 0 to 1.  LSODE mimics Eulerbackward using the options given below. The post shows that LSODE does not do Newton Raphson and just performs only iteration for nonlinear equations.

    restart;

    Digits:=15;

    Digits := 15

    (1)

    eq:=diff(y(t),t)=-y(t);

    eq := diff(y(t), t) = -y(t)

    (2)

    C:=array([0$22]);

    C := Vector[row](22, {(1) = 0, (2) = 0, (3) = 0, (4) = 0, (5) = 0, (6) = 0, (7) = 0, (8) = 0, (9) = 0, (10) = 0, (11) = 0, (12) = 0, (13) = 0, (14) = 0, (15) = 0, (16) = 0, (17) = 0, (18) = 0, (19) = 0, (20) = 0, (21) = 0, (22) = 0})

    (3)

    C[9]:=1;

    C[9] := 1

    (4)

    sol:=dsolve({eq,y(0)=1},type=numeric,method=lsode[backfull],ctrl=C,initstep=0.1,minstep=0.1,abserr=1,relerr=1):

    sol(0.1);

    [t = .1, y(t) = .909090909090834]

    (5)

    subs(diff(y(t),t)=(y1-1)/0.1,y(t)=y1,eq);

    0.1e2*y1-0.1e2 = -y1

    (6)

    fsolve(%,y1=0.5);

    .909090909090909

    (7)

     While for linear it gave the expected result, it gives wrong results for nonlinear problems.

    sol1:=dsolve({eq,y(0)=1},type=numeric):

    sol1(0.1);

    [t = .1, y(t) = .904837355407810]

    (8)

    eq:=diff(y(t),t)=-y(t)^2*exp(-y(t))-10*y(t)*(1+0.01*exp(y(t)));

    eq := diff(y(t), t) = -y(t)^2*exp(-y(t))-10*y(t)*(1+0.1e-1*exp(y(t)))

    (9)

    sol:=dsolve({eq,y(0)=1},type=numeric,method=lsode[backfull],ctrl=C,initstep=0.1,minstep=0.1,abserr=1,relerr=1):

    sol(0.1);

    [t = .1, y(t) = .501579294869466]

    (10)

    subs(diff(y(t),t)=(y1-1)/0.1,y(t)=y1,eq);

    0.1e2*y1-0.1e2 = -y1^2*exp(-y1)-10*y1*(1+0.1e-1*exp(y1))

    (11)

    fsolve(%,y1=1);

    .488691779256025

    (12)

    sol1:=dsolve({eq,y(0)=1},type=numeric):

     the expected answer is correctly obtained with default tolerance as

    sol1(0.1);

    [t = .1, y(t) = .349614721994122]

    (13)

     The results obtained are worse than single iteration using jacobian.

    eq2:=(lhs-rhs)(subs(diff(y(t),t)=(y1-1)/0.1,y(t)=y1,eq));

    eq2 := 0.1e2*y1-0.1e2+y1^2*exp(-y1)+10*y1*(1+0.1e-1*exp(y1))

    (14)

    jac:=unapply(diff(eq2,y1),y1);

    jac := proc (y1) options operator, arrow; 20.+2*y1*exp(-y1)-y1^2*exp(-y1)+.10*exp(y1)+.10*y1*exp(y1) end proc

    (15)

    f:=unapply(eq2,y1);

    f := proc (y1) options operator, arrow; 0.1e2*y1-0.1e2+y1^2*exp(-y1)+10*y1*(1+0.1e-1*exp(y1)) end proc

    (16)

    y0:=1;

    y0 := 1

    (17)

    dy:=-evalf(f(y0)/jac(y0));

    dy := -.508796088545793

    (18)

    ynew:=y0+dy;

    ynew := .491203911454207

    (19)

     Following procedures confirm that it is indeed calling the procedure only at 0 and 0.1, with backdiag giving slightly better results.

    myfun:= proc(x,y) if not type(x,'numeric') or not type(evalf(y),numeric)then 'procname'(x,y);
        else lprint(`Request at x=`,x); -y^2*exp(-y(x))-10*y*(1+0.01*exp(y)); end if; end proc;

    myfun := proc (x, y) if not (type(x, 'numeric') and type(evalf(y), numeric)) then ('procname')(x, y) else lprint(`Request at x=`, x); -y^2*exp(-y(x))-10*y*(1+0.1e-1*exp(y)) end if end proc

    (20)

    sol1:=dsolve({diff(y(x),x)=myfun(x,y(x)),y(0)=1},numeric,method=lsode[backfull],ctrl=C,initstep=0.1,minstep=0.1,abserr=1,relerr=1,known={myfun}):

    sol1(0.1);

    `Request at x=`, 0.

    `Request at x=`, 0.

    `Request at x=`, .1

    `Request at x=`, .1

    [x = .1, y(x) = .501579304183583]

    (21)

    sol2:=dsolve({diff(y(x),x)=myfun(x,y(x)),y(0)=1},numeric,method=lsode[backdiag],ctrl=C,initstep=0.1,minstep=0.1,abserr=1,relerr=1,known={myfun}):

    sol2(0.1);

    `Request at x=`, 0.

    `Request at x=`, 0.

    `Request at x=`, .1

    `Request at x=`, .1

    [x = .1, y(x) = .497831388424072]

    (22)

     

    Download Lsodeanalysistrunc.mws

     

    Next see how dsolve method = mgear works just fine in Maple 6 (gives the expected answer upto 3 Digits accuracy). To run this code you will need Maple 6 or earlier versions. Maple 7 has this algorithm, but I don't know to use it as it is hidden. I would like to get support from other members to get Maplesoft's attention to bring this algorithm back.

    If Mdy/dt = f(y) is solved using mgear algorithm (instead of dy/dt =f ), then one can have a good DAE solver based on this (M being singular). 

     

    restart;

    myfun:= proc(x,y) if not type(x,'numeric') or not type(evalf(y),numeric)then 'procname'(x,y);
        else lprint(`Request at x=`,x); -y^2*exp(-y(x))-10*y*(1+0.01*exp(y)); end if; end proc;

    myfun := proc (x, y) if not (type(x, 'numeric') and type(evalf(y), numeric)) then ('procname')(x, y) else lprint(`Request at x=`, x); -y^2*exp(-y(x))-10*y*(1+0.1e-1*exp(y)) end if end proc

    (1)

    sol2:=dsolve({diff(y(x),x)=myfun(x,y(x)),y(0)=1},{y(x)},numeric,method=mgear[mstepnum],stepsize=0.1,minstep=0.1,errorper=1):

    sol2(0.1);

    `Request at x=`, 0.

    `Request at x=`, .1

    `Request at x=`, .1

    `Request at x=`, .1

    [x = .1, y(x) = .4887165263]

    (2)

     

     

    Download Mgearworks.mws

    This post describes how Maple was used to investigate the Givens rotation matrix, and to answer a simple question about its behavior. The "Givens" part is the medium, but the message is that it really is better to teach, learn, and do mathematics with a tool like Maple.

    The question: If Givens rotations are used to take the vector Y = <5, -2, 1> to Y2 = , about what axis and through what angle will a single rotation accomplish the same thing?

    The Givens matrix G21 takes Y to the vector Y1 =, and the Givens matrix G31 takes Y1 to Y2. Graphing the vectors Y, Y1, and Y2 reveals that Y1 lies in the xz-plane and that Y2 is parallel to the x-axis. (These geometrical observations should have been obvious, but the typical usage of the Givens technique to "zero-out" elements in a vector or matrix obscured this, at least for me.)

    The matrix G = G31 G21 rotates Y directly to Y2; is the axis of rotation the vector W = Y x Y2, and is the angle of rotation the angle  between Y and Y2? To test these hypotheses, I used the RotationMatrix command in the Student LinearAlgebra package to build the corresponding rotation matrix R. But R did not agree with G. I had either the axis or the angle (actually both) incorrect.

    The individual Givens rotation matrices are orthogonal, so G, their product is also orthogonal. It will have 1 as its single real eigenvalue, and the corresponding eigenvector V is actually the direction of the axis of the rotation. The vector W is a multiple of <0, 1, 2> but V = <a, b, 1>, where . Clearly, W  V.

    The rotation matrix that rotates about the axis V through the angle  isn't the matrix G either. The correct angle of rotation about V turns out to be

    the angle between the projections of Y and Y2 onto the plane orthogonal to V. That came as a great surprise, one that required a significant adjustment of my intuition about spatial rotations. So again, the message is that teaching, learning, and doing mathematics is so much more effective and efficient when done with a tool like Maple.

    A discussion of the Givens rotation, and a summary of the actual computations described above are available in the attached worksheet, What Gives with Givens.mw.

         Parallel curves on surfaces. The distance between the points of the curves is measured along the curves of intersection of the surface and perpendicular planes.
         (According to tradition, it also does not make sense.)

    equidistant_curve_surface_MP.mw

     









     

     

    Run the following command in Maple:

    Explore(plot(x^k), k = 1 .. 3);

     

    Once you’ve run the command, move the slider from side to side. Neat, isn’t it?

    With this single line of code, you have built an interactive application that shows the graph of x to the power of various exponent powers.

     

    The Explore command is an application builder. More specifically, the Explore command can programmatically generate interactive content in Maple worksheets.

    Programmatically generated content is inserted into a Maple worksheet by executing Maple commands. For example, when you run the Explore command on an expression, it inserts a collection of input and output controllers, called Embedded Components, into your Maple worksheet. In the preceding example, the Explore command inserts a table containing:

    • a Slider component, which corresponds to the value for the exponent k
    • a Plot component, which shows the graph of x raised to the power for k

    Together these components form an interactive application that can be used to visualize the effect of changing parameter values.

    Explore can be viewed as an easy application creator that generates simple applications with input and output components. Recently added packages for programmatic content generation broaden Maple’s application authoring abilities to form a full development framework for creating customized interactive content in a Maple worksheet. The DocumentTools package contains many of these new tools. Components and Layout are two sub-packages that generate XML using function calls that represents GUI elements, such as embedded components, tables, input, or output. For example, the DocumentTools:-Components:-Plot command creates a new Plot component. These key pieces of functionality provide all of the building blocks needed to create customizable interfaces inside of the Maple worksheet. For me, this new functionality has completely altered my approach to building Maple worksheets and made it much easier to create new applications that can explore hundreds of data sets, visualize mathematical functions, and more.

    I would go so far as to say that the ability to programmatically generate content is one of the most important new sources of functionality over the past few years, and is something that has the potential to significantly alter the way in which we all use Maple. Programmatic content generation allows you to create applications with hundreds of interactive components in a very short period of time when compared to building them manually using embedded components. As an illustration of this, I will show you how I easily created a table with over 180 embedded components—and the logic to control them.

     

    Building an interface for exploring data sets:

    In my previous blog post on working with data sets in Maple, I demonstrated a simple customized interface for exploring country data sets. That post only hinted at the much bigger story of how the Maple programming language was used to author the application. What follows is the method that I used, and a couple of lessons that I learned along the way.

    When I started building an application to explore the country data sets, I began with an approach that I had used to build several MathApps in the past. I started with a blank Maple worksheet and manually added embedded components for controlling input and output. This included checkbox components for each of the world’s countries, drop down boxes for available data sets, and a couple of control buttons for retrieving data to complete my application.

    This manual, piece-by-piece method seemed like the most direct approach, but building my application by hand proved time-consuming, given that I needed to create 180 checkboxes to house all available countries with data. What I really needed was a quicker, more scriptable way to build my interface.

     

    So jumping right into it, you can view the code that I wrote to create the country data application here:PECCode.txt

    Note that you can download a copy of the associated Maple worksheet at the bottom of this page.

     

    I won’t go into too much detail on how to write this code, but the first thing to note is the length of the code; in fewer than 70 lines, this code generates an interface with all of the required underlying code to drive interaction for 180+ checkboxes, 2 buttons and a plot. In fact, if you open up the application, you’ll see that every check box has several lines of code behind it. If you tried to do this by hand, the amount of effort would be multiplied several times over.

    This is really the key benefit to the world of programmatic content generation. You can easily build and rebuild any kind of interactive application that you need using the Maple programming language. The possibilities are endless.

     

    Some tips and tricks:

    There are a few pitfalls to be aware of when you learn to create content with Maple code. One of the first lessons I learned was that it is always important to consider embedded component name collision and name resolution.

    For those that have experimented with embedded components, you may have noticed that Maple’s GUI gives unique names to components that are copied (or added) in a Maple worksheet. For example, the first TextArea component that you add to a worksheet usually has the default name TextArea0. If you next add another TextArea, this new TextArea gets the name TextArea1, so as to not collide with the first component. Similar behaviour can be observed with any other component and even within some component properties such as ‘group’ name.

    Many of the options for commands in the DocumentTools sub-packages can have “action code”, or code that is run when the component is interacted with. When building action code for a generated component, the action code is specified using a long string that encapsulates all of the code. Due to this code being provided as a long string, one trick that I quickly picked up is that it is important to separate out the names for any components into sub-strings inside of a longer cat statement.

    For example, here is a line that is contained within a longer cat statement in the preceding code:

    cat( "DocumentTools:-SetProperty( \"", "ComboBox_0", "\", 'value', \"Internet Users\" );\n" )

    It is necessary to enclose “ComboBox_0” in quotes, as well as to add in escaped quotes in order to have the resulting action code look like (also note the added new line at the end):

    “DocumentTools:-SetProperty( “ComboBox_0”, ‘value’, “Internet Users” );”

    Doing so ensures that when the components are created, the names are not hard-coded to always just look for a given name. This means that the GUI can scrape through the code and update any newly generated components with a new name when needed. This is important if “ComboBox_0” already exists so that the GUI can instead create “ComboBox_1”.

     

    Another challenge for coding applications is adding a component state. One of the most common problems encountered with running any interactive content in Maple is that if state is not persistent, errors can occur when, for example, a play button is clicked but the required procedures have not been run. This is a very challenging problem, which often require solutions like the use of auto-executing start-up code or more involved component programming. Some features in Maple 2016 have started working to address this, but state is still something that usually needs to be considered on an application by application basis.

    In my example, I needed to save the state of a table containing country names so that the interface retains the information for check box state (checked or unchecked) after restart. That is, if I saved the application with two countries selected, I wanted to ensure that when I opened the file again those same two countries would still be selected, both in the interface as well as in the table that is used to generate the plot. Now accomplishing this was a more interesting task: my hack was to insert a DataTable component, which stored my table as an entry of a 1x1 Matrix rtable. Since the rtable that underlies a DataTable is loaded into memory on Maple load, this gave me a way to ensure that the checked country table was loaded on open.

    Here, for example, is the initial creation of this table:

    "if not eval( :-_SelectedCountries )::Matrix then\n",
    "    :-_SelectedCountries := Matrix(1,1,[table([])]):\n",
    "end if;\n",

    For more details, just look for the term: “:-_SelectedCountries” in the preceding code.

    I could easily devote separate posts to discussing in detail each of these two quick tips. Similarly, there’s much more that can be discussed with respect to authoring an interface using programmatic tools from the DocumentTools packages, but I found the best way to learn more about a command is to try it out yourself. Once you do, you’ll find that there are an endless number of combinations for the kinds of interfaces that can be quickly built using programmatic content generation. Several commands in Maple have already started down the path of inserting customized content for their output (see DataSets:-InsertSearchBox and AudioTools:-Preview as a couple of examples) and I can only see this trend growing.

    Finally, I would like to say that getting started with programmatic content generation was intimidating at first, but with a little bit of experimentation, it was a rewarding experience that has changed the way in which I work in Maple. In many cases, I now view output as something that can be customized for any command. More often than not, I turn to commands like ‘Explore’ to create an interface to see how sweeping through parameters effects my results, and any time I want to perform a more complex analysis or visualization for my data, I write code to create interfaces that can more easily be customized and re-used for other applications.

    If you are interested in learning more about this topic, some good examples to get started with are the examples page for programmatic content generation as well as the help pages for the DocumentTools:-Components and DocumentTools:-Layout sub-packages.

    To download a copy of the worksheet used in this post, click here (note that the code can be found in the start-up code of this worksheet): CountryDataPEC.mw To create the datasets interface, simply run the CountrySelection(); command.

    It would be nice if the StringTools package had a function giving the number of primitive words of length n on an alphabet of size a.  The formula for this is well-known but tedious to code.

            General description of the method of solving underdetermined systems of equations. As a particular application of the idea proposed a universal method  kinematic analysis for all kinds of  spatial and planar link mechanisms with any number degrees of freedom.  The method can be used for powerful CAD linkages.   
             http://www.maplesoft.com/applications/view.aspx?SID=154228

           


          Some examples of a much larger number calculated by the proposed method. Examples gathered here not to look for them on the forum and opportunity to demonstrate the method.  Among the examples, I think, there are very complicated.

    https://vk.com/doc242471809_408704758
    https://vk.com/doc242471809_408704572
    https://vk.com/doc242471809_376439263
    https://vk.com/doc242471809_402619761
    https://vk.com/doc242471809_402610228
    https://vk.com/doc242471809_401188803
    https://vk.com/doc242471809_400465891
    https://vk.com/doc242471809_400711315
    https://vk.com/doc242471809_387358164
    https://vk.com/doc242471809_380837279
    https://vk.com/doc242471809_379935473
    https://vk.com/doc242471809_380217387
    https://vk.com/doc242471809_363266817
    https://vk.com/doc242471809_353980472
    https://vk.com/doc242471809_375452868
    https://vk.com/doc242471809_353988163 
    https://vk.com/doc242471809_353986884 
    https://vk.com/doc242471809_353987119
    https://vk.com/doc242471809_324249241
    https://vk.com/doc242471809_324102889
    https://vk.com/doc242471809_322219275
    https://vk.com/doc242471809_437298137
    https://vk.com/doc242471809_437308238
    https://vk.com/doc242471809_437308241
    https://vk.com/doc242471809_437308243
    https://vk.com/doc242471809_437308245
    https://vk.com/doc242471809_437308246
    https://vk.com/doc242471809_437401651
    https://vk.com/doc242471809_437664558

     

     

     

    Points on the coordinate plane

    (Guidance manual for the 6th class)

    Changing the initial coordinates and going through the entire program first, we get a new picture-task

     

        Point_on_co-plane.mws

     

    And Another     Coordinate_plane.mws

     

     

     

    Coordinate axis

    6th class (in Russia)

    Guidance manual for use at lessons (at school)

    Coordinate_line_lesson.mws

    (It is possible bad English)

    Aggregate statistics are calculated by splitting the rows of a DataFrame by each factor in a given column into subsets and computing summary statistics for each of these subsets.

    The following is a short example of how the Aggregate command is used to compute aggregate statistics for a DataFrame with housing data:

    To begin, we construct a DataFrame with housing data: The first column has number of bedrooms, the second has the area in square feet, the third has price.

    bedrooms := <3, 4, 2, 4, 3, 2, 2, 3, 4, 4, 2, 4, 4, 3, 3>:
    area := <1130, 1123, 1049, 1527, 907, 580, 878, 1075,
    1040, 1295, 1100, 995, 908, 853, 856>:
    price := <114700, 125200, 81600, 127400, 88500, 59500, 96500, 113300,
    104400, 136600, 80100, 128000, 115700, 94700, 89400>:
    HouseSalesData := DataFrame([bedrooms, area, price], columns = [Bedrooms, Area, Price]);

    Note that the Bedrooms column has three distinct levels: 2, 3, and 4.

    convert(HouseSalesData[Bedrooms], set);

    The following returns the mean of all other columns for each distinct level in the column, Bedrooms:

    Aggregate(HouseSalesData, Bedrooms);

    Adding the columns option controls which columns are returned.

    Aggregate(HouseSalesData, Bedrooms, columns = [Price])

    Additionally, the tally option returns a tally for each of the levels.

    Aggregate(HouseSalesData, Bedrooms, tally)

    The function option allows for the specification of any command that can be applied to a DataSeries. For example, the Statistics:-Median command computes the median for each of the levels of Bedrooms.

    Aggregate(HouseSalesData, Bedrooms, function = Statistics:-Median);

    By default, Aggregate uses the SplitByColumn command to creates a separate sub-DataFrame for every discrete level in the column given by bycolumn.

    with(Statistics);
    ByRooms := SplitByColumn(HouseSalesData, Bedrooms);

    We can create box plots of the price for subgroups of sales defined by number of bedrooms.

    BoxPlot( map( (m)->m[Price], ByRooms), 
    deciles=false,
    datasetlabels=["2 bdrms", "3 bdrms", "4 bdrms"],
    color=["Red", "Purple", "Blue"]);

     

    I have recorded a short video that walks through this example here: https://youtu.be/e0pqCMyO3ks

    The worksheet for this example can be downloaded here: Aggregate.mw

    We have just released a new version of the Multivariate Calculus Study Guide.  It provides a new section on Vector Calculus, with over 100 additional worked problems, and makes extensive use of Maple’s Clickable Math tools as well as commands.

    Existing study guide customers can get the new content via a free update, available through the Check for Updates system or from our website. See Multivariate Calculus Study Guide 2016 Update for details.

    For more information about this guide, including a full table of contents, visit Multivariate Calculus Study Guide.

     

    eithne

    Hi,
    The latest update to the differential equations Maple libraries (this week, can be downloaded from the Maplesoft R&D webpage for Differential Equations and Mathematical functions) includes new functionality in pdsolve, regarding whether the solution for a PDE or PDE system is or not a general solution.

    In brief, a general solution of a PDE in 1 unknown, that has differential order N, and where the unknown depends on M independent variables, involves N arbitrary functions of M-1 arguments. It is not entirely evident how to extend this definition in the case of a coupled, possibly nonlinear PDE system. However, using differential algebra techniques (automatically used by pdsolve when tackling a PDE system), that extension to define a general solution for a DE system is possible, and also when the system involves ODEs and PDEs, and/or algebraic (that is, non-differential) equations, and/or inequations of the form algebraic*expression <> 0 involving the unknowns, and all of this in the presence of mathematical functions (based on the use of Maple's PDEtools:-dpolyform). This is a very nice case were many different advanced developments come together to naturally solve a problem that otherwise would be rather difficult.

    The issues at the center of this Maple development/post are then:

            a) How do you know whether a PDE or PDE system solution returned is a general solution?

            b) How could you indicate to pdsolve that you are only interested in a general PDE or PDE system solution?

    The answer to a) is now always present in the last line of the userinfo. So input infolevel[pdsolve] := 3 before calling pdsolve, and check what the last line of the userinfo displayed tells.


    The answer to b) is a new option, generalsolution, implemented in pdsolve so that it either returns a general solution or otherwise it returns NULL. If you do not use this new option, then pdsolve works as always: first it tries to compute a general solution and if it fails in doing that it tries to compute a particular solution by separating the variables in different ways, or computing a traveling wave solution or etc. (a number of other well known methods).

     

    The examples that follow are from the help page pdsolve,system, and show both the new userinfo telling whether the solution returned is a general one and the option generalsolution at work.The examples are all of differential equation systems but the same userinfos and generalsolution option work as well in the case of a single PDE.

     

     

    Example 1.

    Solve the determining PDE system for the infinitesimals of the symmetry generator of example 11 from Kamke's book . Tell whether the solution computed is or not a general solution.

    infolevel[pdsolve] := 3

    3

    (1.1)

    The PDE system satisfied by the symmetries of Kamke's ODE example number 11 is

    sys__1 := [diff(xi(x, y), y, y) = 0, diff(eta(x, y), y, y)-2*(diff(xi(x, y), y, x)) = 0, 3*x^r*y^n*(diff(xi(x, y), y))*a+2*(diff(eta(x, y), y, x))-(diff(xi(x, y), x, x)) = 0, 2*(diff(xi(x, y), x))*x^r*y^n*a-x^r*y^n*(diff(eta(x, y), y))*a+eta(x, y)*a*x^r*y^n*n/y+xi(x, y)*a*x^r*r*y^n/x+diff(eta(x, y), x, x) = 0]

    This is a second order linear PDE system, with two unknowns {eta(x, y), xi(x, y)} and four equations. Its general solution is given by the following, where we now can tell that the solution is a general one by reading the last line of the userinfo. Note that because the system is overdetermined, a general solution in this case does not involve any arbitrary function

    sol__1 := pdsolve(sys__1)

    -> Solving ordering for the dependent variables of the PDE system: [xi(x,y), eta(x,y)]

    -> Solving ordering for the independent variables (can be changed using the ivars option): [x, y]
    tackling triangularized subsystem with respect to xi(x,y)
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    tackling triangularized subsystem with respect to eta(x,y)
    <- Returning a *general* solution

     

    {eta(x, y) = -_C1*y*(r+2)/(n-1), xi(x, y) = _C1*x}

    (1.2)

    Next we indicate to pdsolve that n and r are parameters of the problem, and that we want a solution for n <> 1, making more difficult to identify by eye whether the solution returned is or not a general one. Again the last line of the userinfo tells that pdsolve's solution is indeed a general one

    `sys__1.1` := [op(sys__1), n <> 1]

    [diff(diff(xi(x, y), y), y) = 0, diff(diff(eta(x, y), y), y)-2*(diff(diff(xi(x, y), x), y)) = 0, 3*x^r*y^n*(diff(xi(x, y), y))*a+2*(diff(diff(eta(x, y), x), y))-(diff(diff(xi(x, y), x), x)) = 0, 2*(diff(xi(x, y), x))*x^r*y^n*a-x^r*y^n*(diff(eta(x, y), y))*a+eta(x, y)*a*x^r*y^n*n/y+xi(x, y)*a*x^r*r*y^n/x+diff(diff(eta(x, y), x), x) = 0, n <> 1]

    (1.3)

    `sol__1.1` := pdsolve(`sys__1.1`, parameters = {n, r})

    -> Solving ordering for the dependent variables of the PDE system: [r, n, xi(x,y), eta(x,y)]

    -> Solving ordering for the independent variables (can be changed using the ivars option): [x, y]
    tackling triangularized subsystem with respect to r
    tackling triangularized subsystem with respect to n
    tackling triangularized subsystem with respect to xi(x,y)
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    tackling triangularized subsystem with respect to eta(x,y)
    tackling triangularized subsystem with respect to r
    tackling triangularized subsystem with respect to n
    tackling triangularized subsystem with respect to xi(x,y)
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    tackling triangularized subsystem with respect to eta(x,y)
    tackling triangularized subsystem with respect to r
    tackling triangularized subsystem with respect to n
    tackling triangularized subsystem with respect to xi(x,y)
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    tackling triangularized subsystem with respect to eta(x,y)
    tackling triangularized subsystem with respect to n
    tackling triangularized subsystem with respect to xi(x,y)
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    tackling triangularized subsystem with respect to eta(x,y)
    tackling triangularized subsystem with respect to n
    tackling triangularized subsystem with respect to xi(x,y)
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    tackling triangularized subsystem with respect to eta(x,y)
    tackling triangularized subsystem with respect to xi(x,y)
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    tackling triangularized subsystem with respect to eta(x,y)
    <- Returning a *general* solution

     

    {n = 2, r = -5, eta(x, y) = y*(_C1*x+3*_C2), xi(x, y) = x*(_C1*x+_C2)}, {n = 2, r = -20/7, eta(x, y) = -(2/343)*(-6*_C1*x^2-98*x^(8/7)*_C1*a*y-147*_C2*a*x*y)/(x*a), xi(x, y) = _C1*x^(8/7)+_C2*x}, {n = 2, r = -15/7, eta(x, y) = -(1/343)*(-49*_C2*a*x*y-147*x^(6/7)*_C1*a*y+12*_C1*x)/(x*a), xi(x, y) = _C1*x^(6/7)+_C2*x}, {n = 2, r = r, eta(x, y) = -_C1*y*(r+2), xi(x, y) = _C1*x}, {n = -r-3, r = r, eta(x, y) = ((_C1*x+_C2)*r+4*_C1*x+2*_C2)*y/(r+4), xi(x, y) = x*(_C1*x+_C2)}, {n = n, r = r, eta(x, y) = -_C1*y*(r+2)/(n-1), xi(x, y) = _C1*x}

    (1.4)

    map(pdetest, [`sol__1.1`], `sys__1.1`)

    [[0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0], [0, 0, 0, 0]]

    (1.5)

     

    Example 2.

    Compute the solution of the following (linear) overdetermined system involving two PDEs, three unknown functions, one of which depends on 2 variables and the other two depend on only 1 variable.

    sys__2 := [-(diff(F(r, s), r, r))+diff(F(r, s), s, s)+diff(H(r), r)+diff(G(s), s)+s = 0, diff(F(r, s), r, r)+2*(diff(F(r, s), r, s))+diff(F(r, s), s, s)-(diff(H(r), r))+diff(G(s), s)-r = 0]

    The solution for the unknowns G, H, is given by the following expression, were again determining whether this solution, that depends on 3 arbitrary functions, _F1(s), _F2(r), _F3(s-r), is or not a general solution, is non-obvious.

    sol__2 := pdsolve(sys__2)

    -> Solving ordering for the dependent variables of the PDE system: [F(r,s), H(r), G(s)]

    -> Solving ordering for the independent variables (can be changed using the ivars option): [r, s]
    tackling triangularized subsystem with respect to F(r,s)
    First set of solution methods (general or quasi general solution)
    Trying differential factorization for linear PDEs ...
    differential factorization successful.
    First set of solution methods successful
    tackling triangularized subsystem with respect to H(r)
    tackling triangularized subsystem with respect to G(s)
    <- Returning a *general* solution

     

    {F(r, s) = _F1(s)+_F2(r)+_F3(s-r)-(1/12)*r^2*(r-3*s), G(s) = -(diff(_F1(s), s))-(1/4)*s^2+_C2, H(r) = diff(_F2(r), r)-(1/4)*r^2+_C1}

    (1.6)

    pdetest(sol__2, sys__2)

    [0, 0]

    (1.7)

    Example 3.

    Compute the solution of the following nonlinear system, consisting of Burger's equation and a possible potential.

    sys__3 := [diff(u(x, t), t)+2*u(x, t)*(diff(u(x, t), x))-(diff(u(x, t), x, x)) = 0, diff(v(x, t), t) = -v(x, t)*(diff(u(x, t), x))+v(x, t)*u(x, t)^2, diff(v(x, t), x) = -u(x, t)*v(x, t)]

    We see that in this case the solution returned is not a general solution but two particular ones; again the information is in the last line of the userinfo displayed

    sol__3 := pdsolve(sys__3, [u, v])

    -> Solving ordering for the dependent variables of the PDE system: [v(x,t), u(x,t)]

    -> Solving ordering for the independent variables (can be changed using the ivars option): [x, t]
    tackling triangularized subsystem with respect to v(x,t)
    tackling triangularized subsystem with respect to u(x,t)
    First set of solution methods (general or quasi general solution)
    Second set of solution methods (complete solutions)
    Trying methods for second order PDEs
    Third set of solution methods (simple HINTs for separating variables)
    PDE linear in highest derivatives - trying a separation of variables by *
    HINT = *
    Fourth set of solution methods
    Trying methods for second order linear PDEs
    Preparing a solution HINT ...
    Trying HINT = _F1(x)*_F2(t)
    Fourth set of solution methods
    Preparing a solution HINT ...
    Trying HINT = _F1(x)+_F2(t)
    Trying travelling wave solutions as power series in tanh ...
    * Using tau = tanh(t*C[2]+x*C[1]+C[0])
    * Equivalent ODE system: {C[1]^2*(tau^2-1)^2*diff(diff(u(tau),tau),tau)+(2*C[1]^2*(tau^2-1)*tau+2*u(tau)*C[1]*(tau^2-1)+C[2]*(tau^2-1))*diff(u(tau),tau)}
    * Ordering for functions: [u(tau)]
    * Cases for the upper bounds: [[n[1] = 1]]
    * Power series solution [1]: {u(tau) = tau*A[1,1]+A[1,0]}
    * Solution [1] for {A[i, j], C[k]}: [[A[1,1] = 0], [A[1,0] = -1/2*C[2]/C[1], A[1,1] = -C[1]]]
    travelling wave solutions successful.
    tackling triangularized subsystem with respect to v(x,t)
    First set of solution methods (general or quasi general solution)
    Trying differential factorization for linear PDEs ...
    Trying methods for PDEs "missing the dependent variable" ...
    Second set of solution methods (complete solutions)
    Trying methods for second order PDEs
    Third set of solution methods (simple HINTs for separating variables)
    PDE linear in highest derivatives - trying a separation of variables by *
    HINT = *
    Fourth set of solution methods
    Trying methods for second order linear PDEs
    Preparing a solution HINT ...
    Trying HINT = _F1(x)*_F2(t)
    Third set of solution methods successful
    tackling triangularized subsystem with respect to u(x,t)
    <- Returning a solution that *is not the most general one*

     

    {u(x, t) = -_C2*tanh(_C2*x+_C3*t+_C1)-(1/2)*_C3/_C2, v(x, t) = 0}, {u(x, t) = -_c[1]^(1/2)*((exp(_c[1]^(1/2)*x))^2*_C1-_C2)/((exp(_c[1]^(1/2)*x))^2*_C1+_C2), v(x, t) = _C3*exp(_c[1]*t)*_C1*exp(_c[1]^(1/2)*x)+_C3*exp(_c[1]*t)*_C2/exp(_c[1]^(1/2)*x)}

    (1.8)

    pdetest(sol__3, sys__3)

    [0, 0, 0]

    (1.9)

    This example is also good for illustrating the other related new feature: one can now request to pdsolve to only compute a general solution (it will return NULL if it cannot achieve that). Turn OFF userinfos and try with this example

    infolevel[pdsolve] := 1

    This returns NULL:

    pdsolve(sys__3, [u, v], generalsolution)

    Example 4.

    Another where the solution returned is particular, this time for a linear system, conformed by 38 PDEs, also from differential equation symmetry analysis

    sys__4 := [diff(xi[1](x, y, z, t, u), u) = 0, diff(xi[1](x, y, z, t, u), x)-(diff(xi[2](x, y, z, t, u), y)) = 0, diff(xi[2](x, y, z, t, u), u) = 0, -(diff(xi[1](x, y, z, t, u), y))-(diff(xi[2](x, y, z, t, u), x)) = 0, diff(xi[3](x, y, z, t, u), u) = 0, diff(xi[1](x, y, z, t, u), x)-(diff(xi[3](x, y, z, t, u), z)) = 0, -(diff(xi[3](x, y, z, t, u), y))-(diff(xi[2](x, y, z, t, u), z)) = 0, -(diff(xi[1](x, y, z, t, u), z))-(diff(xi[3](x, y, z, t, u), x)) = 0, diff(xi[4](x, y, z, t, u), u) = 0, diff(xi[3](x, y, z, t, u), t)-(diff(xi[4](x, y, z, t, u), z)) = 0, diff(xi[2](x, y, z, t, u), t)-(diff(xi[4](x, y, z, t, u), y)) = 0, diff(xi[1](x, y, z, t, u), t)-(diff(xi[4](x, y, z, t, u), x)) = 0, -(diff(xi[1](x, y, z, t, u), x))+diff(xi[4](x, y, z, t, u), t) = 0, diff(eta[1](x, y, z, t, u), y, y)+diff(eta[1](x, y, z, t, u), z, z)-(diff(eta[1](x, y, z, t, u), t, t))+diff(eta[1](x, y, z, t, u), x, x) = 0, diff(eta[1](x, y, z, t, u), u, u) = 0, diff(eta[1](x, y, z, t, u), u, x)+diff(xi[1](x, y, z, t, u), x, x) = 0, diff(xi[1](x, y, z, t, u), x, y)+diff(eta[1](x, y, z, t, u), u, y) = 0, -(diff(xi[1](x, y, z, t, u), y, y))+diff(eta[1](x, y, z, t, u), u, x) = 0, diff(xi[1](x, y, z, t, u), x, z)+diff(eta[1](x, y, z, t, u), u, z) = 0, diff(xi[1](x, y, z, t, u), y, z) = 0, -(diff(xi[1](x, y, z, t, u), z, z))+diff(eta[1](x, y, z, t, u), u, x) = 0, -(diff(eta[1](x, y, z, t, u), t, u))-(diff(xi[1](x, y, z, t, u), t, x)) = 0, diff(xi[1](x, y, z, t, u), t, y) = 0, diff(xi[1](x, y, z, t, u), t, z) = 0, diff(xi[1](x, y, z, t, u), t, t)+diff(eta[1](x, y, z, t, u), u, x) = 0, -(diff(xi[2](x, y, z, t, u), z, z))+diff(eta[1](x, y, z, t, u), u, y) = 0, diff(xi[2](x, y, z, t, u), t, z) = 0, diff(xi[2](x, y, z, t, u), t, t)+diff(eta[1](x, y, z, t, u), u, y) = 0, diff(xi[3](x, y, z, t, u), t, t)+diff(eta[1](x, y, z, t, u), u, z) = 0, diff(eta[1](x, y, z, t, u), u, x, x) = 0, diff(eta[1](x, y, z, t, u), u, x, y) = 0, diff(eta[1](x, y, z, t, u), u, y, y) = 0, diff(eta[1](x, y, z, t, u), u, x, z) = 0, diff(eta[1](x, y, z, t, u), u, y, z) = 0, diff(eta[1](x, y, z, t, u), u, z, z) = 0, diff(eta[1](x, y, z, t, u), t, u, x) = 0, diff(eta[1](x, y, z, t, u), t, u, y) = 0, diff(eta[1](x, y, z, t, u), t, u, z) = 0]

    There are 38 coupled equations

    nops(sys__4)

    38

    (1.10)

    When requesting a general solution pdsolve returns NULL:

    pdsolve(sys__4, generalsolution)

    A solution that is not a general one, is however computed by default if calling pdsolve without the generalsolution option. In this case again the last line of the userinfo tells that the solution returned is not a general solution

    infolevel[pdsolve] := 3

    3

    (1.11)

    sol__4 := pdsolve(sys__4)

    -> Solving ordering for the dependent variables of the PDE system: [eta[1](x,y,z,t,u), xi[1](x,y,z,t,u), xi[2](x,y,z,t,u), xi[3](x,y,z,t,u), xi[4](x,y,z,t,u)]

    -> Solving ordering for the independent variables (can be changed using the ivars option): [t, x, y, z, u]
    tackling triangularized subsystem with respect to eta[1](x,y,z,t,u)
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    -> Solving ordering for the dependent variables of the PDE system: [_F1(x,y,z,t), _F2(x,y,z,t)]
    -> Solving ordering for the independent variables (can be changed using the ivars option): [t, x, y, z, u]
    tackling triangularized subsystem with respect to _F1(x,y,z,t)
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    -> Solving ordering for the dependent variables of the PDE system: [_F3(x,y,z), _F4(x,y,z)]
    -> Solving ordering for the independent variables (can be changed using the ivars option): [x, y, z, t]
    tackling triangularized subsystem with respect to _F3(x,y,z)
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    tackling triangularized subsystem with respect to _F4(x,y,z)
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    -> Solving ordering for the dependent variables of the PDE system: [_F5(y,z), _F6(y,z)]
    -> Solving ordering for the independent variables (can be changed using the ivars option): [y, z, x]
    tackling triangularized subsystem with respect to _F5(y,z)
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    tackling triangularized subsystem with respect to _F6(y,z)
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    -> Solving ordering for the dependent variables of the PDE system: [_F7(z), _F8(z)]
    -> Solving ordering for the independent variables (can be changed using the ivars option): [z, y]
    tackling triangularized subsystem with respect to _F7(z)
    tackling triangularized subsystem with respect to _F8(z)
    tackling triangularized subsystem with respect to _F2(x,y,z,t)
    First set of solution methods (general or quasi general solution)
    Trying differential factorization for linear PDEs ...
    Trying methods for PDEs "missing the dependent variable" ...
    Second set of solution methods (complete solutions)
    Third set of solution methods (simple HINTs for separating variables)
    PDE linear in highest derivatives - trying a separation of variables by *
    HINT = *
    Fourth set of solution methods
    Preparing a solution HINT ...
    Trying HINT = _F3(x)*_F4(y)*_F5(z)*_F6(t)
    Third set of solution methods successful

    tackling triangularized subsystem with respect to xi[1](x,y,z,t,u)
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    -> Solving ordering for the dependent variables of the PDE system: [_F1(x,z,t), _F2(x,z,t)]
    -> Solving ordering for the independent variables (can be changed using the ivars option): [t, x, z, y]
    tackling triangularized subsystem with respect to _F1(x,z,t)
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    tackling triangularized subsystem with respect to _F2(x,z,t)
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful

    -> Solving ordering for the dependent variables of the PDE system: [_F3(x,t), _F4(x,t)]
    -> Solving ordering for the independent variables (can be changed using the ivars option): [t, x, z]
    tackling triangularized subsystem with respect to _F3(x,t)
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    tackling triangularized subsystem with respect to _F4(x,t)
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    -> Solving ordering for the dependent variables of the PDE system: [_F5(x), _F6(x)]
    -> Solving ordering for the independent variables (can be changed using the ivars option): [x, t]
    tackling triangularized subsystem with respect to _F5(x)
    tackling triangularized subsystem with respect to _F6(x)
    tackling triangularized subsystem with respect to xi[2](x,y,z,t,u)
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    -> Solving ordering for the dependent variables of the PDE system: [_F1(t), _F2(t)]
    -> Solving ordering for the independent variables (can be changed using the ivars option): [t, z]
    tackling triangularized subsystem with respect to _F1(t)
    tackling triangularized subsystem with respect to _F2(t)
    tackling triangularized subsystem with respect to xi[3](x,y,z,t,u)
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    tackling triangularized subsystem with respect to xi[4](x,y,z,t,u)
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    First set of solution methods successful
    <- Returning a solution that *is not the most general one*

     

    {eta[1](x, y, z, t, u) = (_C13*(_C10*(exp(_c[3]^(1/2)*z))^2+_C11)*(_C8*(exp(_c[2]^(1/2)*y))^2+_C9)*(_C6*(exp(_c[1]^(1/2)*x))^2+_C7)*cos((-_c[1]-_c[2]-_c[3])^(1/2)*t)+_C12*(_C10*(exp(_c[3]^(1/2)*z))^2+_C11)*(_C8*(exp(_c[2]^(1/2)*y))^2+_C9)*(_C6*(exp(_c[1]^(1/2)*x))^2+_C7)*sin((-_c[1]-_c[2]-_c[3])^(1/2)*t)+u*exp(_c[1]^(1/2)*x)*exp(_c[2]^(1/2)*y)*exp(_c[3]^(1/2)*z)*(_C1*t+_C2*x+_C3*y+_C4*z+_C5))/(exp(_c[1]^(1/2)*x)*exp(_c[2]^(1/2)*y)*exp(_c[3]^(1/2)*z)), xi[1](x, y, z, t, u) = -(1/2)*_C2*x^2+(1/2)*(-2*_C1*t-2*_C3*y-2*_C4*z+2*_C17)*x+(1/2)*(-t^2+y^2+z^2)*_C2+_C16*t+_C15*z+_C14*y+_C18, xi[2](x, y, z, t, u) = -(1/2)*_C3*y^2+(1/2)*(-2*_C1*t-2*_C2*x-2*_C4*z+2*_C17)*y+(1/2)*(-t^2+x^2+z^2)*_C3+_C20*t+_C19*z-_C14*x+_C21, xi[3](x, y, z, t, u) = -(1/2)*_C4*z^2+(1/2)*(-2*_C1*t-2*_C2*x-2*_C3*y+2*_C17)*z+(1/2)*(-t^2+x^2+y^2)*_C4+_C22*t-_C19*y-_C15*x+_C23, xi[4](x, y, z, t, u) = -(1/2)*_C1*t^2+(1/2)*(-2*_C2*x-2*_C3*y-2*_C4*z+2*_C17)*t+(1/2)*(-x^2-y^2-z^2)*_C1+_C20*y+_C22*z+_C16*x+_C24}

    (1.12)

    pdetest(sol__4, sys__4)

    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

    (1.13)

    Example 5.

    Finally, the new userinfos also tell whether a solution is or not a general solution when working with PDEs that involve anticommutative variables  set using the Physics  package

    with(Physics, Setup)

    [Setup]

    (1.14)

    Set first theta and Q as suffixes for variables of type/anticommutative  (see Setup )

    Setup(anticommutativepre = {Q, theta})

    `* Partial match of  'anticommutativepre' against keyword 'anticommutativeprefix'`

     

    [anticommutativeprefix = {Q, _lambda, theta}]

    (1.15)

    A PDE system example with two unknown anticommutative functions of four variables, two commutative and two anticommutative; to avoid redundant typing in the input that follows and redundant display of information on the screen let's use PDEtools:-diff_table   PDEtools:-declare

    PDEtools:-declare(Q(x, y, theta[1], theta[2]))

    Q(x, y, theta[1], theta[2])*`will now be displayed as`*Q

    (1.16)

    q := PDEtools:-diff_table(Q(x, y, theta[1], theta[2]))

    table( [(  ) = Q(x, y, theta[1], theta[2]) ] )

    (1.17)

    Consider the system formed by these two PDEs (because of the q diff_table just defined, we can enter derivatives directly using the function's name indexed by the differentiation variables)

    pde[1] := q[x, y, theta[1]]+q[x, y, theta[2]]-q[y, theta[1], theta[2]] = 0

    Physics:-diff(diff(diff(Q(x, y, theta[1], theta[2]), x), y), theta[1])+Physics:-diff(diff(diff(Q(x, y, theta[1], theta[2]), x), y), theta[2])-Physics:-diff(Physics:-diff(diff(Q(x, y, theta[1], theta[2]), y), theta[1]), theta[2]) = 0

    (1.18)

    pde[2] := q[theta[1]] = 0

    Physics:-diff(Q(x, y, theta[1], theta[2]), theta[1]) = 0

    (1.19)

    The solution returned for this system is indeed a general solution

    pdsolve([pde[1], pde[2]])

    -> Solving ordering for the dependent variables of the PDE system: [_F4(x,y), _F2(x,y), _F3(x,y)]

    -> Solving ordering for the independent variables (can be changed using the ivars option): [x, y]
    tackling triangularized subsystem with respect to _F4(x,y)
    tackling triangularized subsystem with respect to _F2(x,y)
    tackling triangularized subsystem with respect to _F3(x,y)
    First set of solution methods (general or quasi general solution)
    Trying simple case of a single derivative.
    HINT = _F6(x)+_F5(y)
    Trying HINT = _F6(x)+_F5(y)
    HINT is successful
    First set of solution methods successful
    <- Returning a *general* solution

     

    Q(x, y, theta[1], theta[2]) = _F1(x, y)*_lambda1+(_F6(x)+_F5(y))*theta[2]

    (1.20)

    NULL

    This solution involves an anticommutative constant `_&lambda;2`, analogous to the commutative constants _Cn where n is an integer.

     

    Download PDE_general_solutions.mw

    Edgardo S. Cheb-Terrab
    Physics, Differential Equations and Mathematical Functions, Maplesoft


    Hi

    New developments (after the release of Maple 2016) happened in the project on exact solutions for "Partial Differential Equations & Boundary Conditions". This is work in collaboration with Katherina von Bulow and the improvements are of wide range, representing a noticeable step forward in the capabilities of the Maple system for this kind of problem. As usual, these improvements can be installed in current Maple 2016 by downloading the updated library from the Maplesoft R&D webpage for Differential Equations and Mathematical functions (the update is distributed merged with the updates of the Physics package)

     

    The improvements cover:

     

    • 

    PDE&BC in semi-infinite domains for which a bounded solution is sought

    • 

    PDE & BC problems in bounded spatial domains via eigenfunction (Fourier) expansions

    • 

    Implementation of another algebraic method for tackling linear PDE & BC

    • 

    Improvements in solving PDE & BC solutions by first finding the PDE's general solution.

    • 

    Improvements in solving PDE & BC problems by using a Fourier transform.

    • 

    PDE & BC problems that used to require the option HINT = `+` are now solved automatically

     

    What follows is a set of examples solved now with these new developments, organized in sections according to the kind of problem. Where relevant, the sections include a subsection on "How it works step by step".

    PDE&BC in semi-infinite domains for which a bounded solution is sought can now also be solved via Laplace transforms

     

    Maple is now able to solve more PDE&BC problems via Laplace transforms.

     

    How it works: Laplace transforms act to change derivatives with respect to one of the independent variables of the domain into multiplication operations in the transformed domain. After applying a Laplace transform to the original problem, we can simplify the problem using the transformed BC, then solve the problem in the transformed domain, and finally apply the inverse Laplace transform to arrive at the final solution. It is important to remember to give pdsolve any necessary restrictions on the variables and constants of the problem, by means of the "assuming" command.

     

    A new feature is that we can now tell pdsolve that the dependent variable is bounded, by means of the optional argument HINT = boundedseries.

     

    restart

     

    Consider the problem of a falling cable lying on a table that is suddenly removed (cf. David J. Logan's Applied Partial Differential Equations p.115).

    pde[1] := diff(u(x, t), t, t) = c^2*(diff(u(x, t), x, x))-g
    iv[1] := u(x, 0) = 0, u(0, t) = 0, (D[2](u))(x, 0) = 0

     

    If we ask pdsolve to solve this problem without the condition of boundedness of the solution, we obtain:

    `assuming`([pdsolve([pde[1], iv[1]])], [0 < t, 0 < x, 0 < c])

    u(x, t) = -invlaplace(exp(s*x/c)*_F1(s), s, t)+invlaplace(exp(-s*x/c)*_F1(s), s, t)-(1/2)*g*t^2+invlaplace(exp(s*x/c)/s^3, s, t)*g

    (1.1)

    New: If we now ask for a bounded solution, by means of the option HINT = boundedseries, pdsolve simplifies the problem accordingly.

    ans[1] := `assuming`([pdsolve([pde[1], iv[1]], HINT = boundedseries)], [0 < t, 0 < x, 0 < c])

    u(x, t) = (1/2)*g*(Heaviside(t-x/c)*(c*t-x)^2-c^2*t^2)/c^2

    (1.2)

     

    And we can check this answer against the original problem, if desired:

    `assuming`([pdetest(ans[1], [pde[1], iv[1]])], [0 < t, 0 < x, 0 < c])

    [0, 0, 0, 0]

    (1.3)

    How it works, step by step

     

     Let us see the process this problem undergoes to be solved by pdsolve, step by step.

     

    First, the Laplace transform is applied to the PDE:

    with(inttrans)

    transformed_PDE := laplace((lhs-rhs)(pde[1]), t, s)

    s^2*laplace(u(x, t), t, s)-(D[2](u))(x, 0)-s*u(x, 0)-c^2*(diff(diff(laplace(u(x, t), t, s), x), x))+g/s

    (1.1.1)

    and the result is simplified using the initial conditions:

    simplified_transformed_PDE := eval(transformed_PDE, {iv[1]})

    s^2*laplace(u(x, t), t, s)-c^2*(diff(diff(laplace(u(x, t), t, s), x), x))+g/s

    (1.1.2)

    Next, we call the function "laplace(u(x,t),t,s)" by the new name U:

    eq_U := subs(laplace(u(x, t), t, s) = U(x, s), simplified_transformed_PDE)

    s^2*U(x, s)-c^2*(diff(diff(U(x, s), x), x))+g/s

    (1.1.3)

    And this equation, which is really an ODE, is solved:

    solution_U := dsolve(eq_U, U(x, s))

    U(x, s) = exp(-s*x/c)*_F2(s)+exp(s*x/c)*_F1(s)-g/s^3

    (1.1.4)

    Now, since we want a BOUNDED solution, the term with the positive exponential must be zero, and we are left with:

    bounded_solution_U := subs(coeff(rhs(solution_U), exp(s*x/c)) = 0, solution_U)

    U(x, s) = exp(-s*x/c)*_F2(s)-g/s^3

    (1.1.5)

    Now, the initial solution must also be satisfied. Here it is, in the transformed domain:

    Laplace_BC := laplace(u(0, t), t, s) = 0

    laplace(u(0, t), t, s) = 0

    (1.1.6)

    Or, in the new variable U,

    Laplace_BC_U := U(0, s) = 0

    U(0, s) = 0

    (1.1.7)

    And by applying it to bounded_solution_U, we find the relationship

    simplify(subs(x = 0, rhs(bounded_solution_U))) = 0

    (_F2(s)*s^3-g)/s^3 = 0

    (1.1.8)

    isolate((_F2(s)*s^3-g)/s^3 = 0, indets((_F2(s)*s^3-g)/s^3 = 0, unknown)[1])

    _F2(s) = g/s^3

    (1.1.9)

    so that our solution now becomes

    bounded_solution_U := subs(_F2(s) = g/s^3, bounded_solution_U)

    U(x, s) = exp(-s*x/c)*g/s^3-g/s^3

    (1.1.10)

    to which we now apply the inverse Laplace transform to obtain the solution to the problem:

    `assuming`([u(x, t) = invlaplace(rhs(bounded_solution_U), s, t)], [0 < x, 0 < t, 0 < c])

    u(x, t) = (1/2)*g*(-t^2+Heaviside(t-x/c)*(c*t-x)^2/c^2)

    (1.1.11)

    Four other related examples

     

    A few other examples:

    pde[2] := diff(u(x, t), t, t) = c^2*(diff(u(x, t), x, x))
    iv[2] := u(x, 0) = 0, u(0, t) = g(t), (D[2](u))(x, 0) = 0

    ans[2] := `assuming`([pdsolve([pde[2], iv[2]], HINT = boundedseries)], [0 < t, 0 < x, 0 < c])

    u(x, t) = Heaviside(t-x/c)*g((c*t-x)/c)

    (1.2.1)

    `assuming`([pdetest(ans[2], [pde[2], iv[2]])], [0 < t, 0 < x, 0 < c])

    [0, 0, 0, 0]

    (1.2.2)

    pde[3] := diff(u(x, t), t) = k*(diff(u(x, t), x, x)); iv[3] := u(x, 0) = 0, u(0, t) = 1

    ans[3] := `assuming`([pdsolve([pde[3], iv[3]], HINT = boundedseries)], [0 < t, 0 < x, 0 < k])

    u(x, t) = 1-erf((1/2)*x/(t^(1/2)*k^(1/2)))

    (1.2.3)

    pdetest(ans[3], [pde[3], iv[3][2]])

    [0, 0]

    (1.2.4)

    pde[4] := diff(u(x, t), t) = k*(diff(u(x, t), x, x)); iv[4] := u(x, 0) = mu, u(0, t) = lambda

    ans[4] := `assuming`([pdsolve([pde[4], iv[4]], HINT = boundedseries)], [0 < t, 0 < x, 0 < k])

    u(x, t) = (-lambda+mu)*erf((1/2)*x/(t^(1/2)*k^(1/2)))+lambda

    (1.2.5)

    pdetest(ans[4], [pde[4], iv[4][2]])

    [0, 0]

    (1.2.6)

     

    The following is an example from page 76 in Logan's book:

    pde[5] := diff(u(x, t), t) = diff(u(x, t), x, x)
    iv[5] := u(x, 0) = 0, u(0, t) = f(t)

    ans[5] := `assuming`([pdsolve([pde[5], iv[5]], HINT = boundedseries)], [0 < t, 0 < x])

    u(x, t) = (1/2)*x*(int(f(_U1)*exp(-x^2/(4*t-4*_U1))/(t-_U1)^(3/2), _U1 = 0 .. t))/Pi^(1/2)

    (1.2.7)

    More PDE&BC problems in bounded spatial domains can now be solved via eigenfunction (Fourier) expansions

     

    The code for solving PDE&BC problems in bounded spatial domains has been expanded. The method works by separating the variables by product, so that the problem is transformed into an ODE system (with initial and/or boundary conditions) problem, one of which is a Sturm-Liouville problem (a type of eigenvalue problem) which has infinitely many solutions - hence the infinite series representation of the solutions.

    restart

     

    Here is a simple example for the heat equation:

    pde__6 := diff(u(x, t), t) = k*(diff(u(x, t), x, x)); iv__6 := u(0, t) = 0, u(l, t) = 0

    ans__6 := `assuming`([pdsolve([pde__6, iv__6])], [0 < l])

    u(x, t) = Sum(_C1*sin(_Z1*Pi*x/l)*exp(-k*Pi^2*_Z1^2*t/l^2), _Z1 = 1 .. infinity)

    (2.1)

    pdetest(ans__6, [pde__6, iv__6])

    [0, 0, 0]

    (2.2)

     

    Now, consider the displacements of a string governed by the wave equation, where c is a constant (cf. Logan p.28).

    pde__7 := diff(u(x, t), t, t) = c^2*(diff(u(x, t), x, x))
    iv__7 := u(0, t) = 0, u(l, t) = 0

    ans__7 := `assuming`([pdsolve([pde__7, iv__7])], [0 < l])

    u(x, t) = Sum(sin(_Z2*Pi*x/l)*(sin(c*_Z2*Pi*t/l)*_C1+cos(c*_Z2*Pi*t/l)*_C5), _Z2 = 1 .. infinity)

    (2.3)

    pdetest(ans__7, [pde__7, iv__7])

    [0, 0, 0]

    (2.4)

    Another wave equation problem (cf. Logan p.130):

    pde__8 := diff(u(x, t), t, t)-c^2*(diff(u(x, t), x, x)) = 0; iv__8 := u(0, t) = 0, (D[2](u))(x, 0) = 0, (D[1](u))(l, t) = 0, u(x, 0) = f(x)

    ans__8 := `assuming`([pdsolve([pde__8, iv__8], u(x, t))], [0 <= x, x <= l])

    u(x, t) = Sum(2*(Int(f(x)*sin((1/2)*Pi*(2*_Z3+1)*x/l), x = 0 .. l))*sin((1/2)*Pi*(2*_Z3+1)*x/l)*cos((1/2)*c*Pi*(2*_Z3+1)*t/l)/l, _Z3 = 1 .. infinity)

    (2.5)

    pdetest(ans__8, [pde__8, iv__8[1 .. 3]])

    [0, 0, 0, 0]

    (2.6)

     

    Here is a problem with periodic boundary conditions (cf. Logan p.131). The function u(x, t) stands for the concentration of a chemical dissolved in water within a tubular ring of circumference 2*l. The initial concentration is given by f(x), and the variable x is the arc-length parameter that varies from 0 to 2*l.

    pde__9 := diff(u(x, t), t) = M*(diff(u(x, t), x, x))
    iv__9 := u(0, t) = u(2*l, t), (D[1](u))(0, t) = (D[1](u))(2*l, t), u(x, 0) = f(x)

    ans__9 := `assuming`([pdsolve([pde__9, iv__9], u(x, t))], [0 <= x, x <= 2*l])

    u(x, t) = (1/2)*_C8+Sum(((Int(f(x)*sin(_Z4*Pi*x/l), x = 0 .. 2*l))*sin(_Z4*Pi*x/l)+(Int(f(x)*cos(_Z4*Pi*x/l), x = 0 .. 2*l))*cos(_Z4*Pi*x/l))*exp(-M*Pi^2*_Z4^2*t/l^2)/l, _Z4 = 1 .. infinity)

    (2.7)

    pdetest(ans__9, [pde__9, iv__9[1 .. 2]])

    [0, 0, 0]

    (2.8)

     

    The following problem is for heat flow with both boundaries insulated (cf. Logan p.166, 3rd edition)

    pde__10 := diff(u(x, t), t) = k*(diff(u(x, t), x, x))
    iv__10 := (D[1](u))(0, t) = 0, (D[1](u))(l, t) = 0, u(x, 0) = f(x)

    ans__10 := `assuming`([pdsolve([pde__10, iv__10], u(x, t))], [0 <= x, x <= l])

    u(x, t) = Sum(2*(Int(f(x)*cos(_Z6*Pi*x/l), x = 0 .. l))*cos(_Z6*Pi*x/l)*exp(-k*Pi^2*_Z6^2*t/l^2)/l, _Z6 = 1 .. infinity)

    (2.9)

    pdetest(ans__10, [pde__10, iv__10[1 .. 2]])

    [0, 0, 0]

    (2.10)

     

    This is a problem in a bounded domain with the presence of a source. A source term represents an outside influence in the system and leads to an inhomogeneous PDE (cf. Logan p.149):

    pde__11 := diff(u(x, t), t, t)-c^2*(diff(u(x, t), x, x)) = p(x, t)
    iv__11 := u(0, t) = 0, u(Pi, t) = 0, u(x, 0) = 0, (D[2](u))(x, 0) = 0

    ans__11 := pdsolve([pde__11, iv__11], u(x, t))

    u(x, t) = Int(Sum(2*(Int(p(x, tau1)*sin(_Z7*x), x = 0 .. Pi))*sin(_Z7*x)*sin(c*_Z7*(t-tau1))/(Pi*_Z7*c), _Z7 = 1 .. infinity), tau1 = 0 .. t)

    (2.11)

    Current pdetest is unable to verify that this solution cancells the pde__11 mainly because it currently fails in identifying that there is a fourier expansion in it, but its subroutines for testing the boundary conditions work well with this problem

    pdetest_BC := `pdetest/BC`

    pdetest_BC({ans__11}, [iv__11], [u(x, t)])

    [0, 0, 0, 0]

    (2.12)

     

     

    Consider a heat absorption-radiation problem in the bounded domain 0 <= x and x <= 2, t >= 0:

    pde__12 := diff(u(x, t), t) = diff(u(x, t), x, x)
    iv__12 := u(x, 0) = f(x), (D[1](u))(0, t)+u(0, t) = 0, (D[1](u))(2, t)+u(2, t) = 0

    ans__12 := `assuming`([pdsolve([pde__12, iv__12], u(x, t))], [0 <= x and x <= 2, 0 <= t])

    u(x, t) = (1/2)*_C8+Sum(((Int(f(x)*cos((1/2)*_Z8*Pi*x), x = 0 .. 2))*cos((1/2)*_Z8*Pi*x)+(Int(f(x)*sin((1/2)*_Z8*Pi*x), x = 0 .. 2))*sin((1/2)*_Z8*Pi*x))*exp(-(1/4)*Pi^2*_Z8^2*t), _Z8 = 1 .. infinity)

    (2.13)

    pdetest(ans__12, pde__12)

    0

    (2.14)

    Consider the nonhomogeneous wave equation problem (cf. Logan p.213, 3rd edition):

    pde__13 := diff(u(x, t), t, t) = A*x+diff(u(x, t), x, x)
    iv__13 := u(0, t) = 0, u(1, t) = 0, u(x, 0) = 0, (D[2](u))(x, 0) = 0

    ans__13 := pdsolve([pde__13, iv__13])

    u(x, t) = Int(Sum(2*A*(Int(x*sin(Pi*_Z9*x), x = 0 .. 1))*sin(Pi*_Z9*x)*sin(Pi*_Z9*(t-tau1))/(Pi*_Z9), _Z9 = 1 .. infinity), tau1 = 0 .. t)

    (2.15)

    pdetest_BC({ans__13}, [iv__13], [u(x, t)])

    [0, 0, 0, 0]

    (2.16)

     

    Consider the following Schrödinger equation with zero potential energy (cf. Logan p.30):

    pde__14 := I*h*(diff(f(x, t), t)) = -h^2*(diff(f(x, t), x, x))/(2*m)
    iv__14 := f(0, t) = 0, f(d, t) = 0

    ans__14 := `assuming`([pdsolve([pde__14, iv__14])], [0 < d])

    f(x, t) = Sum(_C1*sin(_Z10*Pi*x/d)*exp(-((1/2)*I)*h*Pi^2*_Z10^2*t/(d^2*m)), _Z10 = 1 .. infinity)

    (2.17)

    pdetest(ans__14, [pde__14, iv__14])

    [0, 0, 0]

    (2.18)

    Another method has been implemented for linear PDE&BC

     

    This method is for problems of the form

     

     "(&PartialD;w)/(&PartialD;t)=M[w]"", w(`x__i`,0) = f(`x__i`)" or

     

    "((&PartialD;)^2w)/((&PartialD;)^( )t^2)="M[w]", w(`x__i`,0) = f(`x__i`), (&PartialD;w)/(&PartialD;t)() ? ()|() ? (t=0) =g(`x__i`)"

     

    where M is an arbitrary linear differential operator of any order which only depends on the spatial variables x__i.

     

    Here are some examples:

    pde__15 := diff(w(x1, x2, x3, t), t)-(diff(w(x1, x2, x3, t), x2, x1))-(diff(w(x1, x2, x3, t), x3, x1))-(diff(w(x1, x2, x3, t), x3, x3))+diff(w(x1, x2, x3, t), x3, x2) = 0
    iv__15 := w(x1, x2, x3, 0) = x1^5*x2*x3NULL

    pdsolve([pde__15, iv__15])

    w(x1, x2, x3, t) = 20*(((1/20)*x2*x3-(1/20)*t)*x1^2+(1/4)*t*(x2+x3)*x1+t^2)*x1^3

    (3.1)

    pdetest(%, [pde__15, iv__15])

    [0, 0]

    (3.2)

     

    Here are two examples for which the derivative with respect to t is of the second order, and two initial conditions are given:

    pde__16 := diff(w(x1, x2, x3, t), t, t) = diff(w(x1, x2, x3, t), x2, x1)+diff(w(x1, x2, x3, t), x3, x1)+diff(w(x1, x2, x3, t), x3, x3)-(diff(w(x1, x2, x3, t), x3, x2))
    iv__16 := w(x1, x2, x3, 0) = x1^3*x2^2+x3, (D[4](w))(x1, x2, x3, 0) = -x2*x3+x1

    pdsolve([pde__16, iv__16])

    w(x1, x2, x3, t) = x1^3*x2^2+x3-t*x2*x3+t*x1+3*t^2*x2*x1^2+(1/6)*t^3+(1/2)*t^4*x1

    (3.3)

    pdetest(%, [pde__16, iv__16])

    [0, 0, 0]

    (3.4)

    pde__17 := diff(w(x1, x2, x3, t), t, t) = diff(w(x1, x2, x3, t), x2, x1)+diff(w(x1, x2, x3, t), x3, x1)+diff(w(x1, x2, x3, t), x3, x3)-(diff(w(x1, x2, x3, t), x3, x2))
    iv__17 := w(x1, x2, x3, 0) = x1^3*x3^2+sin(x1), (D[4](w))(x1, x2, x3, 0) = cos(x1)-x2*x3

    pdsolve([pde__17, iv__17])

    w(x1, x2, x3, t) = (1/2)*t^4*x1+t^2*x1^3+3*t^2*x1^2*x3+x1^3*x3^2+(1/6)*t^3-t*x2*x3+cos(x1)*t+sin(x1)

    (3.5)

    pdetest(%, [pde__17, iv__17])

    [0, 0, 0]

    (3.6)

    More PDE&BC problems are now solved via first finding the PDE's general solution.

     

    The following are examples of PDE&BC problems for which pdsolve is successful in first calculating the PDE's general solution, and then fitting the initial or boundary condition to it.

    pde__18 := diff(u(x, y), x, x)+diff(u(x, y), y, y) = 0
    iv__18 := u(0, y) = sin(y)/y

    If we ask pdsolve to solve the problem, we get:

    ans__18 := pdsolve([pde__18, iv__18])

    u(x, y) = (sin(-y+I*x)+_F2(y-I*x)*(y-I*x)+(-y+I*x)*_F2(y+I*x))/(-y+I*x)

    (4.1)

    and we can check this answer by using pdetest:

    pdetest(ans__18, [pde__18, iv__18])

    [0, 0]

    (4.2)

    How it works, step by step:

     

    The general solution for just the PDE is:

    gensol := pdsolve(pde__18)

    u(x, y) = _F1(y-I*x)+_F2(y+I*x)

    (4.1.1)

    Substituting in the condition iv__18, we get:

    u(0, y) = sin(y)/y

    (4.1.2)

    gensol_with_condition := eval(rhs(gensol), x = 0) = rhs(iv__18)

    _F1(y)+_F2(y) = sin(y)/y

    (4.1.3)

    We then isolate one of the functions above (we can choose either one, in this case), convert it into a function operator, and then apply it to gensol

    _F1 = unapply(solve(_F1(y)+_F2(y) = sin(y)/y, _F1(y)), y)

    _F1 = (proc (y) options operator, arrow; (-_F2(y)*y+sin(y))/y end proc)

    (4.1.4)

    eval(gensol, _F1 = (proc (y) options operator, arrow; (-_F2(y)*y+sin(y))/y end proc))

    u(x, y) = (-_F2(y-I*x)*(y-I*x)-sin(-y+I*x))/(y-I*x)+_F2(y+I*x)

    (4.1.5)

     

     

    Three other related examples

     

    pde__19 := diff(u(x, y), x, x)+(1/2)*(diff(u(x, y), y, y)) = 0
    iv__19 := u(0, y) = sin(y)/y

    pdsolve([pde__19, iv__19])

    u(x, y) = (2*sin(-y+((1/2)*I)*2^(1/2)*x)+(-I*2^(1/2)*x+2*y)*_F2(y-((1/2)*I)*2^(1/2)*x)+(I*2^(1/2)*x-2*y)*_F2(y+((1/2)*I)*2^(1/2)*x))/(I*2^(1/2)*x-2*y)

    (4.2.1)

    pdetest(%, [pde__19, iv__19])

    [0, 0]

    (4.2.2)

    pde__20 := diff(u(x, y), x, x)+(1/2)*(diff(u(x, y), y, y)) = 0
    iv__20 := u(x, 0) = sin(x)/x

    pdsolve([pde__20, iv__20])

    u(x, y) = (sinh((1/2)*(I*2^(1/2)*x-2*y)*2^(1/2))*2^(1/2)-(I*2^(1/2)*x-2*y)*(_F2(-y+((1/2)*I)*2^(1/2)*x)-_F2(y+((1/2)*I)*2^(1/2)*x)))/(I*2^(1/2)*x-2*y)

    (4.2.3)

    pdetest(%, [pde__20, iv__20])

    [0, 0]

    (4.2.4)

    pde__21 := diff(u(r, t), r, r)+(diff(u(r, t), r))/r+(diff(u(r, t), t, t))/r^2 = 0
    iv__21 := u(3, t) = sin(6*t)

    ans__21 := pdsolve([pde__21, iv__21])

    u(r, t) = -_F2(-(2*I)*ln(3)+I*ln(r)+t)+sin(-(6*I)*ln(3)+(6*I)*ln(r)+6*t)+_F2(-I*ln(r)+t)

    (4.2.5)

    pdetest(ans__21, [pde__21, iv__21])

    [0, 0]

    (4.2.6)

    More PDE&BC problems are now solved by using a Fourier transform.

     

    restart

    Consider the following problem with an initial condition:

    pde__22 := diff(u(x, t), t) = diff(u(x, t), x, x)+m
    iv__22 := u(x, 0) = sin(x)

     

    pdsolve can solve this problem directly:

    ans__22 := pdsolve([pde__22, iv__22])

    u(x, t) = sin(x)*exp(-t)+m*t

    (5.1)

    And we can check this answer against the original problem, if desired:

    pdetest(ans__22, [pde__22, iv__22])

    [0, 0]

    (5.2)

    How it works, step by step

     

    Similarly to the Laplace transform method, we start the solution process by first applying the Fourier transform to the PDE:

    with(inttrans)

    transformed_PDE := fourier((lhs-rhs)(pde__22) = 0, x, s)

    -2*m*Pi*Dirac(s)+s^2*fourier(u(x, t), x, s)+diff(fourier(u(x, t), x, s), t) = 0

    (5.1.1)

    Next, we call the function "fourier(u(x,t),x,s1)" by the new name U:

    transformed_PDE_U := subs(fourier(u(x, t), x, s) = U(t, s), transformed_PDE)

    -2*m*Pi*Dirac(s)+s^2*U(t, s)+diff(U(t, s), t) = 0

    (5.1.2)

    And this equation, which is really an ODE, is solved:

    solution_U := dsolve(transformed_PDE_U, U(t, s))

    U(t, s) = (2*m*Pi*Dirac(s)*t+_F1(s))*exp(-s^2*t)

    (5.1.3)

    Now, we apply the Fourier transform to the initial condition iv__22:

    u(x, 0) = sin(x)

    (5.1.4)

    transformed_IC := fourier(iv__22, x, s)

    fourier(u(x, 0), x, s) = I*Pi*(Dirac(s+1)-Dirac(s-1))

    (5.1.5)

    Or, in the new variable U,

    trasnformed_IC_U := U(0, s) = rhs(transformed_IC)

    U(0, s) = I*Pi*(Dirac(s+1)-Dirac(s-1))

    (5.1.6)

    Now, we evaluate solution_U at t = 0:

    solution_U_at_IC := eval(solution_U, t = 0)

    U(0, s) = _F1(s)

    (5.1.7)

    and substitute the transformed initial condition into it:

    eval(solution_U_at_IC, {trasnformed_IC_U})

    I*Pi*(Dirac(s+1)-Dirac(s-1)) = _F1(s)

    (5.1.8)

    Putting this into our solution_U, we get

    eval(solution_U, {(rhs = lhs)(I*Pi*(Dirac(s+1)-Dirac(s-1)) = _F1(s))})

    U(t, s) = (2*m*Pi*Dirac(s)*t+I*Pi*(Dirac(s+1)-Dirac(s-1)))*exp(-s^2*t)

    (5.1.9)

    Finally, we apply the inverse Fourier transformation to this,

    solution := u(x, t) = invfourier(rhs(U(t, s) = (2*m*Pi*Dirac(s)*t+I*Pi*(Dirac(s+1)-Dirac(s-1)))*exp(-s^2*t)), s, x)

    u(x, t) = sin(x)*exp(-t)+m*t

    (5.1.10)

    PDE&BC problems that used to require the option HINT = `+` to be solved are now solved automatically

     

    The following two PDE&BC problems used to require the option HINT = `+` in order to be solved. This is now done automatically within pdsolve.

    pde__23 := diff(u(r, t), r, r)+(diff(u(r, t), r))/r+(diff(u(r, t), t, t))/r^2 = 0
    iv__23 := u(1, t) = 0, u(2, t) = 5

    ans__23 := pdsolve([pde__23, iv__23])

    u(r, t) = 5*ln(r)/ln(2)

    (6.1)

    pdetest(ans__23, [pde__23, iv__23])

    [0, 0, 0]

    (6.2)

    pde__24 := diff(u(x, y), y, y)+diff(u(x, y), x, x) = 6*x-6*y

    iv__24 := u(x, 0) = x^3+11*x+1, u(x, 2) = x^3+11*x-7, u(0, y) = -y^3+1, u(4, y) = -y^3+109

    ans__24 := pdsolve([pde__24, iv__24])

    u(x, y) = x^3-y^3+11*x+1

    (6.3)

    pdetest(ans__24, [pde__24, iv__24])

    [0, 0, 0, 0, 0]

    (6.4)

    ``



    Download PDE_and_BC_update.mw

    Edgardo S. Cheb-Terrab
    Physics, Differential Equations and Mathematical Functions, Maplesoft

    We've added a collection of thermal engineering applications to the Application Center. You could think of it as an e-book.

    This collection has a few features that I think are pretty neat

    • The applications are collected together in a Workbook; a single file gives you access to 30 applications
    • You can navigate the contents using the Navigator or a hyperlinked table of contents
    • You can change working fluids and operating conditions, while still using accurate thermophysical data

    If you don't have Maple 2016, you can view and navigate the applications (and interactive with a few) using the free Player.

    The collection includes these applications.

    • Psychrometric Modeling
      • Swamp Cooler
      • Adiabatic Mixing of Air
      • Human Comfort Zone
      • Dew Point and Wet Bulb Temperature
      • Interactive Psychrometric Chart
    • Thermodynamic Cycles
      • Ideal Brayton Cycle
      • Optimize a Rankine Cycle
      • Efficiency of a Rankine Cycle
      • Turbine Analysis
      • Organic Rankine Cycle
      • Isothermal Compression of Methane
      • Adiabatic Compression of Methane
    • Refrigeration
      • COP of a Refrigeration Cycle
      • Flow Through an Expansion Valve
      • Food Refrigeration
      • Rate of Refrigerant Boiling
      • Refrigeration Cycle Analysis 1
      • Refrigeration Cycle Analysis 2
    • Miscellaneous
      • Measurement Error in a Manometer
      • Particle Falling Through Air
      • Saturation Temperature of Fluids
      • Water Fountain
      • Water in Piston
    • Heat Transfer
      • Dittus-Boelter Correlation
      • Double Pipe Heat Exchanger
      • Energy Needed to Vaporize Ethanol
      • Heat Transfer Coefficient Across a Flat Plate
    • Vapor-Liquid Equilibria
      • Water-Ethanol

    I have a few ideas for more themed Maple application collections. Data analysis, anyone?

    First 8 9 10 11 12 13 14 Last Page 10 of 250